Can a quantum state over time resemble a quantum state at a single - - PowerPoint PPT Presentation

Can a quantum state over time resemble a quantum state at a single time?

Chris Heunen

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History

Barbados, 2011

• J. Barrett

Information processing in generalized probabilistic theories, Physical Review A 75(3):032304, 2007 2 / 20

History

Barbados, 2011

• J. Barrett

Information processing in generalized probabilistic theories, Physical Review A 75(3):032304, 2007

Oxford, 2012

• C. Heunen and C. Horsman

Matrix multiplication is determined by orthogonality and trace, Linear Algebra and its Applications 439(12):4130–4134, 2013. 2 / 20

Prehistory

Barbados, 2011

• J. Barrett

Information processing in generalized probabilistic theories, Physical Review A 75(3):032304, 2007

Oxford, 2012

• C. Heunen and C. Horsman

Matrix multiplication is determined by orthogonality and trace, Linear Algebra and its Applications 439(12):4130–4134, 2013.

London, 2013

• M. Pusey

Is quantum steering spooky? PhD, Imperial College London, 2013. 2 / 20

Prehistory

Barbados, 2011

• J. Barrett

Information processing in generalized probabilistic theories, Physical Review A 75(3):032304, 2007

Oxford, 2012

• C. Heunen and C. Horsman

Matrix multiplication is determined by orthogonality and trace, Linear Algebra and its Applications 439(12):4130–4134, 2013.

London, 2013

• M. Pusey

Is quantum steering spooky? PhD, Imperial College London, 2013.

Kitchener, 2014

• M. Leifer and R. Spekkens

Towards a formulation of quantum theory as a causality neutral theory of Bayesian inference , Physical Review A 88(5):052130, 2013 2 / 20

Prehistory

Barbados, 2011 Oxford, 2013

• J. Barrett

Information processing in generalized probabilistic theories, Physical Review A 75(3):032304, 2007

Oxford, 2012 Durham, 2015

• C. Heunen and C. D. Horsman

Matrix multiplication is determined by orthogonality and trace, Linear Algebra and its Applications 439(12):4130–4134, 2013.

London, 2013 Kitchener, 2014

• M. Pusey

Is quantum steering spooky? PhD, Imperial College London, 2013.

Kitchener, 2014

• M. Leifer and R. Spekkens

Towards a formulation of quantum theory as a causality neutral theory of Bayesian inference , Physical Review A 88(5):052130, 2013 2 / 20

Idea

◮ ‘problem of time’ in quantum gravity

asymmetry between space and time in quantum theory

◮ one solution: remove asymmetry

quantum states across time as well as space

◮ many approaches to states over time

path integrals, consistent histories, multi-time states

◮ but: all depend on spatio-temporal relationships

classical probability theory does not

◮ can it be done?

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Outline

◮ Three proposals ◮ Four axioms ◮ The theorem ◮ Its proof ◮ What does it mean?

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States across space

A B first region second region HA HB ρA ρB

States across space

A B first region second region HA HB ρA ρB composite system HA ⊗ HB ρA ⊗ ρB

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States across time

A B first time evolution second time

States across time

A B first time evolution second time HA HB

States across time

A B first time evolution second time HA HB ρA EAB ρB

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States across time

A B first time evolution second time HA HB ρA EAB ρB Composite system AB, operator ρAB on HAB = HA ⊗ HB No restrictions on ρAB, but fully defined by ρA, ρB, and EAB ρAB = f(ρB, EAB, ρA)

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States across time

A B first time evolution second time HA HB ρA EAB ρB Composite system AB, operator ρAB on HAB = HA ⊗ HB No restrictions on ρAB, but fully defined by ρA and EAB ρAB = f(EAB, ρA) What form could f take?

6 / 20

States across time

A B first time evolution second time HA HB ρA EAB ρB Composite system AB, operator ρAB on HAB = HA ⊗ HB No restrictions on ρAB, but fully defined by ρA and EAB ρAB = EAB ⋆ ρA What form could star product ⋆: B(HAB) × B(HAB) → B(HAB) take? channel state EAB =

ij EAB|ij|A ⊗ |ji|B

satisfying ρB = TrA(EABρA)

6 / 20

Leifer-Spekkens

“Quantum theory is like Bayesian inference”

Towards a formulation of quantum theory as a causality neutral theory of Bayesian inference Physical Review A 88(5):052130, 2013

channel state conditional probabilities EAB P(B | A) TrB(EAB) = 1

• B P(B | A) = 1

ρB = TrA(EAB ρA)

• A P(B | A) P(A)

? P(AB) = P(B | A) P(A)

7 / 20

Leifer-Spekkens

“Quantum theory is like Bayesian inference”

Towards a formulation of quantum theory as a causality neutral theory of Bayesian inference Physical Review A 88(5):052130, 2013

channel state conditional probabilities EAB P(B | A) TrB(EAB) = 1

• B P(B | A) = 1

ρB = TrA(EAB ρA)

• A P(B | A) P(A)

? P(AB) = P(B | A) P(A) ? = ρ(LS)

AB

= EAB ⋆LS ρA = √ρA EAB √ρA

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Fitzsimons-Jones-Vedral

“Use pseudo-density matrices”

Quantum correlations which imply causation Scientific Reports 5:18281, 2015

For (multi-)qubit systems A, B:

• 1. Measure σi ∈ {1, σx, σy, σz} on A
• 2. Evolve according to channel EAB
• 3. Measure σj on B

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Fitzsimons-Jones-Vedral

“Use pseudo-density matrices”

Quantum correlations which imply causation Scientific Reports 5:18281, 2015

For (multi-)qubit systems A, B:

• 1. Measure σi ∈ {1, σx, σy, σz} on A
• 2. Evolve according to channel EAB
• 3. Measure σj on B

ρ(FJV)

AB

= 1 4 3

• i=1

σi ⊗ σjσi ⊗ σj

• 8 / 20

Fitzsimons-Jones-Vedral

“Use pseudo-density matrices”

Quantum correlations which imply causation Scientific Reports 5:18281, 2015

For (multi-)qubit systems A, B:

• 1. Measure σi ∈ {1, σx, σy, σz} on A
• 2. Evolve according to channel EAB
• 3. Measure σj on B

ρ(FJV)

AB

= 1 4 3

• i=1

σi ⊗ σjσi ⊗ σj

• = EAB ⋆FJV ρA = 1

2(ρAEAB + EABρA)

8 / 20

Discrete Wigner functions

“Use quasi-probabilities on discrete representation”

On the quantum correction for thermodynamic equilibrium Physical Review 40:749-759, 1932

• 1. Pick phase-point operator basis {KA

i } for HA

Tr(KA

i KA j ) = δij dim(HA) and i KA i = dim(HA)

• 2. Write ρA as quasi-probability function rA : {i} → [−1, 1]
• 3. Write EAB as conditional quasi-probability rB|A : {(i, j)} → [−1, 1]
• 4. Define ρAB by rAB(ij) = rB|A(j | i)rA(i)

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Discrete Wigner functions

“Use quasi-probabilities on discrete representation”

On the quantum correction for thermodynamic equilibrium Physical Review 40:749-759, 1932

• 1. Pick phase-point operator basis {KA

i } for HA

Tr(KA

i KA j ) = δij dim(HA) and i KA i = dim(HA)

• 2. Write ρA as quasi-probability function rA : {i} → [−1, 1]
• 3. Write EAB as conditional quasi-probability rB|A : {(i, j)} → [−1, 1]
• 4. Define ρAB by rAB(ij) = rB|A(j | i)rA(i)

ρ(W)

AB =

• ij

rB|A(j | i)rA(i)KA

i ⊗ KB j

9 / 20

Discrete Wigner functions

“Use quasi-probabilities on discrete representation”

On the quantum correction for thermodynamic equilibrium Physical Review 40:749-759, 1932

• 1. Pick phase-point operator basis {KA

i } for HA

Tr(KA

i KA j ) = δij dim(HA) and i KA i = dim(HA)

• 2. Write ρA as quasi-probability function rA : {i} → [−1, 1]
• 3. Write EAB as conditional quasi-probability rB|A : {(i, j)} → [−1, 1]
• 4. Define ρAB by rAB(ij) = rB|A(j | i)rA(i)

ρ(W)

AB =

• ij

rB|A(j | i)rA(i)KA

i ⊗ KB j

= EAB ⋆W ρA =

• ij

TrAB(EABKA

i ⊗ KB j ) TrA(ρAKA i )KA i ⊗ KB j

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Axiom 1: preservation of probabilistic mixtures

If A conditioned on fair classical coin ρA,x=h = |00|, ρA,x=t = |11| and channel is identity EAB = |φ+φ+|TB then should have composite state be mixture EAB ⋆ 1 2ρA,x=h + 1 2ρA,x=t

• = 1

2

• EAB ⋆ ρA,x=h
• + 1

2 (EAB ⋆ ρA,x=t)

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Axiom 1: preservation of probabilistic mixtures

If A conditioned on fair classical coin ρA,x=h = |00|, ρA,x=t = |11| and channel is identity EAB = |φ+φ+|TB then should have composite state be mixture EAB ⋆ 1 2ρA,x=h + 1 2ρA,x=t

• = 1

2

• EAB ⋆ ρA,x=h
• + 1

2 (EAB ⋆ ρA,x=t) Axiom 1: convex-bilinearity

• px + (1 − p)y
• ⋆ z = p(x ⋆ z) + (1 − p)(y ⋆ z)

x ⋆

• py + (1 − p)z) = p(x ⋆ y) + (1 − p)(x ⋆ z)

for all operators x, y, z and probabilities p ∈ [0, 1]

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Axiom 2: preservation of classical limit

If channel completely dephasing EAB =

ii|ρ|i j p(j | i) |jj|

and input state diagonal ρA =

i p(i) |ii|

then should reproduce joint classical probabilities EAB ⋆ ρA =

• i

p(i) |ii| ⊗

• i

p(j | i)|jj| = EABρA

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Axiom 2: preservation of classical limit

If channel completely dephasing EAB =

ii|ρ|i j p(j | i) |jj|

and input state diagonal ρA =

i p(i) |ii|

then should reproduce joint classical probabilities EAB ⋆ ρA =

• i

p(i) |ii| ⊗

• i

p(j | i)|jj| = EABρA Axiom 2: product on commuting pairs [x, y] = 0 = ⇒ x ⋆ y = xy for all operators x, y

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Axiom 3: preservation of marginals

Joint state ρAB should reproduce marginal states ρA and ρB TrB ρAB = ρA TrA ρAB = ρB TrB(EAB ⋆ ρA) = TrB(EABρA) TrA(EAB ⋆ ρA) = TrA(EABρA)

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Axiom 3: preservation of marginals

Joint state ρAB should reproduce marginal states ρA and ρB TrB ρAB = ρA TrA ρAB = ρB TrB(EAB ⋆ ρA) = TrB(EABρA) TrA(EAB ⋆ ρA) = TrA(EABρA) Axiom 3: product when traced Tr(x ⋆ y) = Tr(xy) for all operators x, y

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Axiom 4: preservation of compositionality

A B C EAB EBC ρA ρB ρC form EAB,C, get ρABC = EAB,C ⋆ ρAB

• r

form EA,BC, get ρABC = EA,BC ⋆ ρA

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Axiom 4: preservation of compositionality

A B C EAB EBC ρA ρB ρC form EBC, get ρABC = EBC ⋆ (EAB ⋆ ρA)

• r

form EA,BC, get ρABC = EA,BC ⋆ ρA

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Axiom 4: preservation of compositionality

A B C EAB EBC ρA ρB ρC form EBC, get ρABC = EBC ⋆ (EAB ⋆ ρA)

• r

form EA,BC, get ρABC = (EBC ⋆ EAB) ⋆ ρA

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Axiom 4: preservation of compositionality

A B C EAB EBC ρA ρB ρC form EBC, get ρABC = EBC ⋆ (EAB ⋆ ρA)

• r

form EA,BC, get ρABC = (EBC ⋆ EAB) ⋆ ρA Axiom 4: associativity (x ⋆ y) ⋆ z = x ⋆ (y ⋆ z) for all operators x, y, z

13 / 20

Three proposals, four axioms

convex- product on product bilinear commuting pairs when traced associative LS ×

• ×

FJV W Recall x ⋆LS y = y1/2 x y1/2 if x = |φ+φ+|TB and y = |00|, z = |11|, and p = 1/2, then x ⋆ (py + (1 − p)z) = p(x ⋆ y) + (1 − p)(x ⋆ z)

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Three proposals, four axioms

convex- product on product bilinear commuting pairs when traced associative LS ×

• ×

FJV

• ×

W Recall x ⋆FJV y = 1

2(xy + yx)

in general (x ⋆ y) ⋆ z = x ⋆ (y ⋆ z)

14 / 20

Three proposals, four axioms

convex- product on product bilinear commuting pairs when traced associative LS ×

• ×

FJV

• ×*

W Recall x ⋆FJV y = 1

2(xy + yx)

in general (x ⋆ y) ⋆ z = x ⋆ (y ⋆ z) But for single qubits: EBC ⋆ (EAB ⋆ ρA) = (EBC ⋆ EAB) ⋆ ρA because ρA commutes with EBC

14 / 20

Three proposals, four axioms

convex- product on product bilinear commuting pairs when traced associative LS ×

• ×

FJV

• ×*

W

• ×
• Recall x ⋆W y =

ij TrAB(EABKA i ⊗ KB j ) TrA(ρAkA i )KA i ⊗ KB j

Unless quasi-probabilities are non-negative, [x, y] = 0 ⇒ x ⋆ y = xy because joint state not diagonal in eigenbasis

• f input state and channel state even if they commute

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Main result

Star theorem: Let Mn be the complex vector space of n-by-n matrices. The only functions ⋆: Mn × Mn → Mn satisfying the four axioms are x ⋆ y = xy and x ⋆ y = yx.

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Main result

Star theorem: Let Mn be the complex vector space of n-by-n matrices. The only functions ⋆: Mn × Mn → Mn satisfying the four axioms are x ⋆ y = xy and x ⋆ y = yx. An operator ρ: HA ⊗ HB → HA ⊗ HB is locally positive when it gives nonnegative probabilities on local measurements: ab|ρ|ab ≥ 0.

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Main result

Star theorem: Let Mn be the complex vector space of n-by-n matrices. The only functions ⋆: Mn × Mn → Mn satisfying the four axioms are x ⋆ y = xy and x ⋆ y = yx. An operator ρ: HA ⊗ HB → HA ⊗ HB is locally positive when it gives nonnegative probabilities on local measurements: ab|ρ|ab ≥ 0. Reduction lemma: Let LPn be the set of locally positive n-by-n matrices. Any function ⋆: LPn × LPn → LPn satisfying the axioms extends to a function ⋆: Mn × Mn → Mn satisfying the axioms.

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The proof: reduction lemma

First extend from LPn to set Hn of hermitian matrices:

◮ conic combination of xi ∈ Hn is i rixi for ri ∈ [0, ∞) ◮ define ⋆ on conic hull of LPn by

(

• i

rixi) ⋆ (

• j

sjyj) =

• ij

risj(xi ⋆ yj)

◮ unique convex-bilinear extension ⋆: Hn × Hn → Hn

(because x ⋆ 0 = 0 = 0 ⋆ x) in fact R-bilinear

◮ other three axioms preserved by extension

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The proof: reduction lemma

Next extend from Hn to Mn:

◮ polarize x ∈ Mn as x = xh + xa

for xh = 1

2(x + x†) ∈ Hn and xa = 1 2(x − x†) ◮ define ⋆ on Mn by

x ⋆ y = xh ⋆ yh − i(xh ⋆ iya) − i(ixa ⋆ yh) − (ixa ⋆ iya)

◮ unique R-bilinear extension ⋆: Mn × Mn → Mn

in fact C-bilinear

◮ other three axioms preserved by extension

(product on commuting pairs uses Fuglede’s theorem: xy = yx and yy† = y†y imply xy† = y†x)

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The proof

◮ Product on commuting pairs implies:

x ⋆ y = 0 if xy = yx = 0, xx = x, yy = y, for rank one x, y ∈ Mn

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The proof

On bilinear maps determined by rank one idempotents Linear Algebra and its Applications 432:738743, 2010

◮ Product on commuting pairs implies:

x ⋆ y = 0 if xy = yx = 0, xx = x, yy = y, for rank one x, y ∈ Mn

◮ Additionally C-bilinearity implies:

x ⋆ y = xy + g(xy − yx) for some linear map g: Mn → Mn and all x, y ∈ Mn

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The proof

On bilinear maps determined by rank one idempotents Linear Algebra and its Applications 432:738743, 2010

◮ Product on commuting pairs implies:

x ⋆ y = 0 if xy = yx = 0, xx = x, yy = y, for rank one x, y ∈ Mn

◮ Additionally C-bilinearity implies:

x ⋆ y = xy + g(xy − yx) for some linear map g: Mn → Mn and all x, y ∈ Mn

◮ Additionally product when traced implies:

g preserves traceless matrices

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The proof: linear algebra

Finally, associativity means for standard matrix units eij − δdeδafg(ecb) + δdeeab ⋆ g(ecf) − δcfeab ⋆ g(eed) = − δafδbcg(eed) + δbcg(ead) ⋆ eef − δadg(ecb) ⋆ eef

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The proof: linear algebra

Finally, associativity means for standard matrix units eij − δdeδafg(ecb) + δdeeab ⋆ g(ecf) − δcfeab ⋆ g(eed) = − δafδbcg(eed) + δbcg(ead) ⋆ eef − δadg(ecb) ⋆ eef Write g(eij) = n

k,l=1 Gkl,ij elk for entries Gkl,ij ∈ C.

Write g(ij)kl for Gkl,ij, and g(ii − jj)kl for Gkl,ii − Gkl,jj.

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The proof: linear algebra

Finally, associativity means for standard matrix units eij − δdeδafg(ecb) + δdeeab ⋆ g(ecf) − δcfeab ⋆ g(eed) = − δafδbcg(eed) + δbcg(ead) ⋆ eef − δadg(ecb) ⋆ eef Write g(eij) = n

k,l=1 Gkl,ij elk for entries Gkl,ij ∈ C.

Write g(ij)kl for Gkl,ij, and g(ii − jj)kl for Gkl,ii − Gkl,jj. For distinct i, j, k, l, the following all vanish: g(ij)kl g(ii − jj)kl g(ij)jk g(ij)kj g(ij)ik g(ij)ki g(ij)kk g(ii − jj)kk g(ii − jj)ik g(ii − jj)ki g(ii − jj)kj g(ii − jj)jk g(ij)jj g(ij)ii g(ij)ij g(ii − jj)ij

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The proof: linear algebra

Finally, associativity means for standard matrix units eij − δdeδafg(ecb) + δdeeab ⋆ g(ecf) − δcfeab ⋆ g(eed) = − δafδbcg(eed) + δbcg(ead) ⋆ eef − δadg(ecb) ⋆ eef Write g(eij) = n

k,l=1 Gkl,ij elk for entries Gkl,ij ∈ C.

Write g(ij)kl for Gkl,ij, and g(ii − jj)kl for Gkl,ii − Gkl,jj. For distinct i, j, k, l, the following all vanish: g(ij)kl g(ii − jj)kl g(ij)jk g(ij)kj g(ij)ik g(ij)ki g(ij)kk g(ii − jj)kk g(ii − jj)ik g(ii − jj)ki g(ii − jj)kj g(ii − jj)jk g(ij)jj g(ij)ii g(ij)ij g(ii − jj)ij More fiddling: there is z ∈ Mn with g(ii) = λeii + z, and λ ∈ {0, −1} with g(ii − jj)ii = −g(ii − jj)jj = g(ij)ji = λ for i = j

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The proof: linear algebra

Finally, associativity means for standard matrix units eij − δdeδafg(ecb) + δdeeab ⋆ g(ecf) − δcfeab ⋆ g(eed) = − δafδbcg(eed) + δbcg(ead) ⋆ eef − δadg(ecb) ⋆ eef Write g(eij) = n

k,l=1 Gkl,ij elk for entries Gkl,ij ∈ C.

Write g(ij)kl for Gkl,ij, and g(ii − jj)kl for Gkl,ii − Gkl,jj. For distinct i, j, k, l, the following all vanish: g(ij)kl g(ii − jj)kl g(ij)jk g(ij)kj g(ij)ik g(ij)ki g(ij)kk g(ii − jj)kk g(ii − jj)ik g(ii − jj)ki g(ii − jj)kj g(ii − jj)jk g(ij)jj g(ij)ii g(ij)ij g(ii − jj)ij More fiddling: there is z ∈ Mn with g(ii) = λeii + z, and λ ∈ {0, −1} with g(ii − jj)ii = −g(ii − jj)jj = g(ij)ji = λ for i = j Conclusion: x ⋆ y = xy + λ(xy − yx)

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Conclusion

◮ Quantum probabilistic reasoning cannot be Bayesian

quantum probabilities compose differently over space and time

◮ States over time not attractive for quantum gravity

need quantum way to discern space and time not in relativity

◮ Larger state spaces don’t help

⋆: X × X′ → X′′ for X, X′ ⊆ Mm and X′′ ⊆ Mn

◮ Star theorem holds without product when traced for qubits

proof by brute force, might hold in any dimension

◮ Dropping associativity most likely way out

joint FJV state is matrix product without imaginary eigenvalues

◮ Other possibility: operational point of view

define joint state via experimental outcomes at different times

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FJV as a star product

Recall ρ(FJV)

AB

= 1 4 3

• i=1

σi ⊗ σjσi ⊗ σj

• Compute

σi ⊗ σj = Tr(EAB((ρi+

A − ρi− A ) ⊗ σj)

= Tr(EAB(1 2(σiρA + ρAσi) ⊗ σj) = 1 2 Tr(ρAEAB(σi ⊗ σj) + EABρA(σi ⊗ σj)) = Tr(ρ(FJV)

AB

(σi ⊗ σj)) So ρ(FJV)

AB

= 1 2(ρA EAB + EAB ρA)

20 / 20